Respuesta :
Answer
given,
mass of car 1, m₁ = 2200 Kg
Speed of car in east, v₁ = 10 m/s
Mass of the car 2, m₂ = 3000 Kg
Speed of car 2 in north, v₂ = ?
Speed of the system after collision, v = 5.23 m/s
Angle of velocity after. θ = 36.0°
Velocity Horizontal component
v_x = v cos θ
v_x = 5.23 cos 36°
v_x = 4.23 m/s
Vertical velocity component
v_y = v sin θ
v_y = 5.23 sin 36°
v_y = 3.074 m/s
a) writing conservation of equation in north direction
m₂v₂ = (m₁+m₂)v_y
3000 x v₂ = (3000+2200) x 3.074
v₂ = 5.33 m/s
b) Initial Kinetic energy
[tex]KE_1 = \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2[/tex]
[tex]KE_1 = \dfrac{1}{2}\times 2200\times 10^2 + \dfrac{1}{2}\times 3000\times 5.33^2[/tex]
[tex]KE_1 = 152613.35\ J[/tex]
Final Kinetic energy
[tex]KE_2 = \dfrac{1}{2}(m_1+m_2)v^2[/tex]
[tex]KE_2 = \dfrac{1}{2}\times 5200\times 5.23^2[/tex]
[tex]KE_2 =71117.54\ J[/tex]
Hence, Decrease in Kinetic energy
[tex]\Delta KE = KE_1-KE_2[/tex]
[tex]\Delta KE = 152613.35-71117.54[/tex]
[tex]\Delta KE = 81495.81\ J[/tex]
Decrease in Kinetic energy after collision is equal to [tex]\Delta KE = 81495.81\ J[/tex]
